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Translation: Photograph of a rectangular pyramid, its vertices and sides
If the lengths of the edges are x, y and z units, then the total surface area is 2*(xy + yz + zx).
hexagon because the word hexa means 5 and it includes the edges and verticies too y* * * * *What utter rubbish.Hexa is a prefix for 6, not 5.A hexagon is a 2-dimensional figure, not a 3d shape.The correct answer is a quadrilateral based pyramid. Rectangular or square based pyramids are the better known versions of this shape.
Properties of a rectangular prism: * A rectangular prism has a total of 6 surfaces. * Because it is rectangular it will have 4 equal longer edges and 8 equal shorter edges, and so * It will have 4 rectangular faces and 2 square faces, therefore * Total surface area = 2 x square (end ) surface + 4 x rectangular (side) surfaces If we let x = shorter edges (breadth) & y = longer edges (length), and, as area = length x breadth, then * each end (a square surface) will have an area of x2 * each side (rectangular surface) will have an area of xy Thus: Total surface area of the rectangular prism = 2 x end area + 4 x side area = 2x2 + 4xy = 2x(x + 2y) Hope this nswers your question and explains how we arrive at the formula for calculating the total surface area of a rectangular prism.
A ball has no faces, that is y it has no edges and no
a grid in a rectangular shape with a x axis and a y axis.
(0,0)
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You also need an equation for y in order to convert to rectangular form.
A point's y coordinate is its vertical position, or how high or low it is.
Let the width of the rectangular prism be y Therefore we can write all the measurements of the rectangular prism in terms of y. width = y length = 2y height = y + 4 Finding the volume of the rectangular prism y x 2y x (y + 4) = 48 2y2(y+4) = 48 y2(y+4) = 24 From this equation, we can see that y must equal 2 (22 x 6 = 24) Therefore the width of the rectangular prism is 2m. Note that in this question, when we got to the equation y2(y+4) = 24, we used 'guess and check' to find that y was. We purposely avoided expanding the bracket as this would have left us with a cubic to solve, which is quite difficult to solve analytically unless you already know one of its roots.